The permutations and combinations of abcd taken 3 at a time are respectively.

Step 1. Calculate the permutations.
When n objects are taken into consideration, the number of permutations is given by,
$P(n,r)=\frac{n!}{(n-r)!}$
We have to find permutations of $abcd$ taken $3$ at a time.
$\therefore P(4,3)=\frac{4!}{(4-3)!}\Rightarrow =\frac{4\times 3\times 2\times 1}{1}\Rightarrow =24$
Step 2. Figure out the combination..
The number of combinations of $n$ objects taken $r$ at a time is given by,
$C(n,r)=\frac{n!}{r!(n-r)!}$
We have to find combinations of $abcd$ taken $3$ at a time.
$\therefore C(4,3)=\frac{4!}{3!(4-3)!}=\frac{4\times 3\times 2\times 1}{3\times 2\times 1\times 1}=4$
Hence, the permutation and combination are $24$ and $4$respectively.